二分法subsequence

subsequence题目网址

算法：当数据量很大适宜采用该方法。采用二分法查找时，数据需是排好序的。 基本思想：假设数据是按升序排序的，对于给定值x，从序列的中间位置开始比较，如果当前位置值等于x，则查找成功；若x小于当前位置值，则在数列的前半段 中查找；若x大于当前位置值则在数列的后半段中继续查找，直到找到为止。

题目描述

A sequence of N positive integers (10 < N < 100 000), each of them less than or equal 10000, and a positive integer S (S < 100 000 000) are given. Write a program to find the minimal length of the subsequence of consecutive elements of the sequence, the sum of which is greater than or equal to S.

input

The first line is the number of test cases. For each test case the program has to read the numbers N and S, separated by an interval, from the first line. The numbers of the sequence are given in the second line of the test case, separated by intervals. The input will finish with the end of file.

output

For each the case the program has to print the result on separate line of the output file.if no answer, print 0.

sample input

2
10 15
5 1 3 5 10 7 4 9 2 8
5 11
1 2 3 4 5

output

2

3

题意：有T组测试数据，有n个正整数组成一个序列，给定整数S，求长度最短的连续序列，使它们的和大于或等于S。

用二分法。输入数时直接让数组等于你刚刚输入的数与前一个数之和。

#include
#include
using namespace std;
int a[100005];
int main()
{
int t;
int temp;
scanf("%d",&t);
while(t--)
{
int n,s;
int left,right;
int ans=0;
scanf("%d%d",&n,&s);
a[0]=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
   a[i]=a[i]+a[i-1];
}
       if(a[n]        printf("0\n");
       else
       {
        left=1;right=n;
        int mid;
        while(left         {
        mid=(left+right)/2;
        int flag=0;
        for(int i=mid;i<=n;i++)
        {
        if((a[i]-a[i-mid])>=s)
        {
        flag=1;
        break;
}
}
if(flag==1)
right=mid;
else
left=mid;
}
printf("%d\n",right);
  }
}

return 0;
}